B How Do Dedekind Cuts Affect Addition and Negative Numbers?

[bhobba]

This is a really basic stupid question but has me beat. I am almost certainly missing something but for the life of me cat see what. If A and B are are the lower sets of two Dedekind cuts addition is defined as the sum of the elements (r+s) with r in A and s in B. Is that correct or have I missed something? What I cant see is why if x is a negative number x + 0 = x. As I said I am likely missing something basic but for the life of me can't see what it is. Working on a paper defining the numbers but using a different approach based on hyperrationals and hyperreals. This came up while writing it but like I said has me beat. I will probably feel terrible when told where I went wrong about such a basic question.

Thanks
Bill

 

[jedishrfu]

I don’t know which is worse death by a Dedekind cut or by a thousand other cuts.

Perhaps, @fresh_42 can answer your question best.

In the meantime, I found this reference on math.stackexchange

https://math.stackexchange.com/questions/1259453/definition-of-dedekind-cut-addition

They add an irrational with its negative and then ask if 0 is a member.

and this homework chapter

http://homepages.math.uic.edu/~groves/teaching/2008-9/313/09-313Hw2Sols.pdf

 

[bhobba]

Thanks. I will quote what one reference says and spell out my issue. By the definition of addition of Dedekind cuts above we have f(a)+f(b) is the Dedekind cut determined by the set{x+y|x∈A,y∈B}. Well, the set of rational numbers of the form x+y where x<a and y<b and x,y∈Q is exactly the set of rational numbers less than a+b. Therefore, {x+y|x∈A,y∈B}= C, so f(a)+f(b)=f(a+b).

Take x as -1. The Dedekind cut is A1 = {Q| Q < -1}. Take y as 0. The Dedekind cut is A2 = {Q| Q < 0}. Let the element from A1 be -1.000001. Let the element from A2 be -.000001. Then -1.000001 - .000001 = -1.000002 < -1.

I see my error. Dumb mental arithmetic.

Teaches me a lesson. Don't do stuff in your head - when necessary write it out. Yes and I do feel stupid, but will leave it up anyway so others can learn from my mistake.

Thanks
Bill

 

[fresh_42]

As a service to other readers and for us to have common ground, I'll quote the definitions in my book.

$A\subset \mathbb{Q}$ is called a (Dedekind) cut if

1. $\emptyset \neq A \neq \mathbb{Q}$
2. $\alpha \in A\, , \,\beta \ge \alpha \Longrightarrow \beta \in A$
3. $A$ does not contain a minimal element.

This is a really basic stupid question but has me beat. I am almost certainly missing something but for the life of me cat see what. If A and B are are the lower sets of two Dedekind cuts addition is defined as the sum of the elements (r+s) with r in A and s in B. Is that correct or have I missed something?

My book says: If $A$ and $B$ are cuts then $A+B=\{r+s | r \in A, s \in B\},$ i.e. the author (Christian Blatter) doesn't refer explicitly to "lower sets" but to cuts instead.

What I cant see is why if x is a negative number x + 0 = x.

I can't see where negative should matter. The neutral cut is defined as $S_0=\{\xi\in \mathbb{Q}\,|\,\xi>0\}.$

As I said I am likely missing something basic but for the life of me can't see what it is. Working on a paper defining the numbers but using a different approach based on hyperrationals and hyperreals. This came up while writing it but like I said has me beat. I will probably feel terrible when told where I went wrong about such a basic question.

Thanks
Bill

The proof for $A+S_0=A$ goes as follows:

We get from the second condition that $A+S_0\subseteq A.$ For a given $\alpha\in A$ there is always an $\alpha' < \alpha$ which is still in $A$ by the third condition. Hence
$$
\alpha=\alpha' +(\alpha - \alpha') \in A+ S_0
$$
so $A\subseteq A+S_0$ and $A=A+S_0.$

 

[MidgetDwarf]

if you want a good source to have that walks you through, look at Bloch : Real Numbers and Real Analysis.

 

[wrobel]

Perhaps the Dedekind cuts is not the best way to introduce reals. There is a general way to complete a metric space.

Df: We shall say that $\{x_n\}\subset\mathbb{Q}$ is a Cauchy sequence iff for any (rational) $\varepsilon>0$ there is a number $N$ such that
$$n,m>N\Longrightarrow |x_n-x_m|<\varepsilon.$$

Let $M$ be a set of Cauchy sequences.

Th: The following relation in $M$ is an equivalence relation:
$$\{x_n\}\sim\{y_k\}\Longleftrightarrow |x_n-y_n|\to 0.$$

Df: $\mathbb{R}:=M/\sim$

UPDATE
Let $p:M\to\mathbb{R}$ be the projection. Define a set $U_r\subset \mathbb{R},\quad r>0$ as follows.
$$p(\{x_k\})\in U_r$$ iff there exist $K,\varepsilon>0$ such that
$$k>K\Longrightarrow |x_k|<r-\varepsilon.$$ The sets $U_r$ form a base of neighbourhoods of the origin.

etc

 

[bhobba]

Thanks, Wrrebel.

I am exploring all this in an insights article on what numbers are.

My error was dumb - relying on mental arithmetic instead of doing it on paper. I am so embarrassed.

While writing the article, which is nearly finished, I discovered something very interesting - it's all interrelated to a consistency condition needed to define the hyperrationals well. Each hyperrational must be <,=, > a rational. That forces the finite hyperrationals to be a Cauchy Sequence and a Dedekind cut.

It was a surprising result that emerged while writing the article.

I hope people find reading as interesting as I did in writing it.

Thanks
Bill